Steel Ductility

7/21/2019 Steel Ductility

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M

Ductility: Quantitative Descriptions

M p

yield max


M

M p

yield max

Ductility Factor: µ = max

yield


M

M p

yield max

p

Plastic Rotation Angle: p = max - yield


M

M p

yield max

p

Rotation Capacity: R = p

yield

= µ - 1


M

M p

yield max

Ductility: Ductility: ductility factor µ

plastic rotation angle p

rotation capacity R

etc.

Based on:

yield

max

Ductility: Difficulties with Quantitative Descriptions

M

Consider a more realistic load - deformation response......



M

What isyield

yield

?

M

What isyield

yield

?

M

What is max

max

?

M ma x M

What is max

max

?

0.8 M max

M

What ismax

max

?

M p


M

Many definitions of ducti lity

Many definitions of yield and max




Ductility under cyclic loading.....

∆

-40000

-30000

-20000

-10000

0

10000

20000

30000

40000

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08

Rotation Angle (rad)

B e n d i n g M o m e n t ( k i p - i n c h e s )


Ductility under cyclic loading.....

How should ductili ty be measured ??

What is Ductility ?

Duct i l i ty = inelast ic deformat ion capaci ty

Many ways to quantify ducti lity

When quantifying ducti lity.......

Clearly define measure of ductil ity

Clearly define yield and max

Use consistent definitions when describi ng

ductili ty demand and ductil ity supply

F

∆

F

∆

F

∆

F

∆

Ductility = Yielding

How is ductility developed in steel structures ?

Loss of load carrying

capability:

Instability

Fracture

Why is Ductility Important?

Permits redistribution of internal stresses and

forces

Increases strength of members, connections and

structures

Permits design based on simple equilibrium mo dels

Results in more robust structures

Provides warning of failure

Permits structur e to survive severe earthquake

loading

Why Ductility ?


forces


structures

Permits design based on simple equilibrium mod els



Permits structure to su rvive severe earthquake

loading



Example: Plate wi th ho le subjected to tens ion

PP

P 1/2" x 6"L1" dia. hole

ε

50 ksi X

Material " A"

ε

50 ksi

Material " B"

6"

Example:

PP50 ksi

max = 2.57 avg

50 ksi = 2.57 xP

2.5 in2

Pmax = 49 k

ε

50 ksi X

Material " A"

max

Example:

PP

ε

50 ksi

Material " B"

50 ksi

Example:

PP

ε

50 ksi

Material " B"

50 ksi

50 ksi =P

2.5 in2

Pmax = 125 k

PP

ε

50 ksi X

Material " A"

ε

50 ksi

Material " B"

Pmax = 49k Pmax = 125k

Example: Flexural Capacity

ε

50 ksi X

Material " A"

ε

50 ksi

Material " B"

MM

4"

12"



ε

50 ksi X

Material " A"

4"

12"

max = 50 ksi

ksi 50 S

M ma x =

S = 96 in 3

M max = 96 in 3 x 50 ksi = 4800 k-in

4"

12"

max = 50 ksi

ε

50 ksi

Material " B"

4"

12"

50 ksi

ε

50 ksi

Material " B"

50 ksi

ksi 50 Z

M ma x =

Z = 144 in 3

M max = 144 in 3 x 50 ksi = 7200 k-in

ε

50 ksi X

Material " A"

ε

50 ksi

Material " B"

MM

4"

12"

Mmax = 4800 k-in Mmax = 7200 k-in

Example: Beam Capacity

L = 30 ft.

w

M500 k-ft.

Beam "A"

500 k-ft.

M

Beam "B"


w

M500 k-ft.

Beam "A"

M

500 k-ft

250 k-ft

8 wL

2

ft k

2

750 8

wL −=

w max = 6.67 k / ft.




w

M

500 k-ft

250 k-ft

500 k-ft

8

wL2

ft k

2

1000 8

wL −=

w max = 8.89 k / ft.

500 k-ft.

M

Beam "B"

L = 30 ft.

w

M500 k-ft.

Beam "A"

500 k-ft.

M

Beam "B"

w max = 6.67 k / ft . w max = 8.89 k / ft.

Why Ductility ?


forces


structures





loading

Lower Bou nd Theorem of Plast ic Analysis

A li mit load based on an in ternal s tres s or for ce

distribution that satisfies:

1. Equi librium

2. Material Strength Limits for Ductile Response

( ≤F y , M≤ M p , P ≤ P y , etc)

is less than or equal to the true limit load.

Lower bound theorem only applicable for ductile structures

Implications of the lower bound theorem ............Implications of the lower bound theorem ............

For a structure made of duc tile materials and

components:

Designs satisfying equilibr ium and material

strength lim its are safe.

As a desig ner , as long as we sati sfy equili br ium

(i.e. provide a load path), a ductile s tructur e will

redistribut e internal stresses and forces so as to

find the available load path.

Example of lower bound theorem: Beam Capacity

L = 30 ft.

w

M p

= 500 k-ft.

M

Ductile flexural behavior

What is the load capacity for this beam ?? w max = 8.89 k / ft.



L = 30 ft.

w

What is the load capacity for this beam ??

By lower bound theorem:

Choose any moment diagram in equilibrium

with the applied load.

The moment cannot exceed M p at any point

along the beam.

The resulting load capacity "w " will be less than

or equal to the true load capacity.

L = 30 ft.

w

8

wL2

Moment diagram in equilibrium with

applied load "w"

Possible lower bound solut ions......

L = 30 ft.

w

8

wL 2

M

500 k-ft

ft k

2

500

8

wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )

L = 30 ft.

w

8

wL2

M

500 k-ft

ft k 2

500 8

wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )

L = 30 ft.

w

8

wL2 M

500 k-ft

ft k

2

750 8

wL −= w = 6.67 k / ft. (≤ 8.89 k / ft. )

250 k-ft

L = 30 ft.

w

8

wL2

M

500 k-ft

ft k

2

1000 8

wL −= w = 8.89 k / ft. (= true w max )

500 k-ft



Examples of lower bound theorem

Flexural capacity of steel section:

F y

F y

≤F y

C

T

Equilibrium: C = T

d

M n = C * d = Z F y

Examples of lower bound theorem

Flexural capacity of a composite section:

F y

steel ≤F y

Equilibrium: C = T

M n = C * d

0.85 f c '

C

Td

conc ≤0.85 f c '

Why Ductility ?


forces


structures





loading

Why Ductility ?


forces


structures





loading



Ground

Acceleration

Building:

Mass = m

Building

Acceleration

F = ma

Earthquake Forces

on Buildings:

Inertia Force Due to

Accelerating Mass

Conventional Building Code Philosophy forConventional Building Code Philosophy for

EarthquakeEarthquake--Resistant DesignResistant Design

Objective: Prevent collapse in the extreme

earthquake likely to occur at a

building site.

Objectives are not to:

- limit damage

- maintain function

- provide for easy repair

To Survive Strong Earthquake

without Collapse:

Design for Ductile Behavior Design for Ductile Behavior

H

HDuctility = Inelastic Deformation

HH

Required Strength

MAX

Helastic

3/4 *Helastic

1/2 *Helastic

1/4 *Helastic

Available Ductility



How Do We Achieve Ductility inSteel Structures ?

Achieving Ducti le Response....

Ductile Limit States Must Precede Brittle Limit States

ExampleExample

double angle tension member

gusset plate

PP

Du ct il e L im it State: Gro ss -s ec ti on y ie ld in g of t en si on membe r

B ri tt le Li mi t States : Ne t- sec ti on f ra cture o f te ns io n member

Block-shear fracture of tension member

Net-section fracture of gusset plate

Block-shear fracture of gusset plate

Bolt shear fracture

Plate bearing failure in double angles or guss et


PP

Example: Gross-section y ie ld ing of tens ion

member must precede net section

fracture of tension member

Gross-section yield: P yield = Ag F y

Net-section fracture: P fracture = Ae F u


PP

P yield ≤P fracture

Ag F y ≤Ae F u

u

y

g

e

F

F

A

A≥

The required strength for brittle limit

states is defined by the capacity of the

ductile element

u

y

F

F = yield ratio

Steels with a low yield ratio are

preferable for ductile behavior


PP

Example: Gross-section yield ing of tension member

must precede bolt shear fracture

Gross-section yield: P yield = Ag F y

Bolt shear fracture: P bolt-fracture = n b n s Ab F v F v=

0.4 F u-bolt -N

0.5 F u-bolt -X




PP

P yield ≤P bolt-fracture The required strength for brittle limit

states is defined by the capacity of the

ductile element

The ductile element must be the

weakest element in the load path


PP

Example: Bolts: 3 - 3/4" A325-X double shear

Ab = 0.44 in 2 F v = 0.5 x 120 ksi = 60 ksi

P bolt-fracture = 3 x 0.44 in 2 x 60 ksi x 2 = 158k

Ang les: 2L 4 x 4 x 1/4 A36

Ag = 3.87 in 2

P yield = 3.87 in2 x 36 ksi = 139k


PP

P yield ≤P bolt-fracture

P yield = 139k P bolt-fracture = 158 k OK

What if the actual yield stress for the A36 angles is

greater than 36 ksi?

Say, for example, the actual yield stress for the A36angle is 54 ksi.


PP


P yield = 3.87 in2 x 54 ksi = 209k

P bolt-fracture = 158 k


Bolt fracture will occur before yield of angles non-ductile behavior


PP

P yield ≤P brittle

Stronger is not b etter in the ductile element

(Ductile element must be weakest element in the load path)

For ductile response: must consider material overstrength in

ductile element


PP

P yield ≤P brittle

The required strength for brittle limit

states is defined by the expected

capacity of the ductile element (not

minimum specified capacity)

P yield = Ag R y F y R y F y = expected yield stress of angles



Achieving Ductile Response....

Ductile Limit States Must Precede Brittle Limit States

Define the required strength for brittle limit states based

on the expected yield capacity for ductile element

The ductile element must be the weakest in the load path

Unanticipated over strength in the ductile element can

lead to non-ductile behavior.

Steels with a low value of yi eld ratio, F y / F u are preferable

for ductil e elements


Connection respons e is generally non-duc tile.....

Connections should be stronger than connected

members




Be cautious of high-strength steels

Ref: Salmon and Johnson -Steel

Structu res: Design and Behavior

General Trends:

As F y

Elongation (material

ductility)

F y / F u




Be cautious of high-strength steels

High strength steels are generally less ductil e

(lower elongations) and generally have a higher

yield ratio.

High strength steels are generally undesirable

for ductile elements


Use Sections wi th Low Width-Thickness Ratios and

Adeq uate Lateral Br acing

M

M p

Increasing b / t

Effect of Local Buckling on Flexural Strength and Ductility

M

Mr

M o m e n t C a p a c i t y

λp λr Width-Thickness Ratio (b/t)

Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

D u c t i l i t y

Mr

M

o m e n t C a p a c i t y


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

D u c t i l i t y

Slender Element Sections

Mr

M

o m e n t C a p a c i t y


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

D u c t i l i t y

Noncompact Sections



Mr



Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

D u c t i l i t y

Compact Sections

Mr



Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

D u c t i l i t y

Seismically

Compact Sections

Local buckling of noncompact and slender element sections

Local buckling of a seismically compact moment frame beam.....

-5000

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

5000

-0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05

Drift Angle (radian)

B e n d i n g M

o m e n t ( k N - m )

RBS Connection

Mp

Mp

Local buckling of a seismically compact EBF link.....



-200

-150

-100

-50

0

50

100

150

200

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08

Link Rotation, (rad)

L i n k

S h e a r F o r c e

( k i p s )

Effect of Local Buckling on Ductility

For ductile flexural response:

Use compact or seismically compact sections

Example: W-Shape

b f

t f

h

t w

y

s

f

f

F

E 38 .0

t 2

b ≤

Beam Flanges

Beam Web

y

s

w F

E 45 .2

t

h ≤

Compact:

y

s

f

f

F

E 30 .0

t 2

b ≤

Seismically Compact:

Compact:

Seismically Compact:

y

s

w F

E 76 .3

t

h ≤

Lateral Torsional Bu cklingLateral Torsional Bu ckling

Lateral torsional buckling

controlled by:

y

b

r

L

Lb = distance between beam lateral braces

r y = weak axis radius of gyration

L b L b

Beam lateral braces

M

M p

Increasing Lb / r y

Effect of Lateral Torsional Buckling on Flexural Strength and DuEffect of Lateral Torsional Buckling on Flexural Strength and Ductility:ctility:

M



Effect of Lateral Buckling on Ductility

For ductile flexural response:

Use lateral bracing based on plastic design

requirements or seismic design requirements

Plastic Design: y

y 2

1

b r

F

E

M

M 076 .0 12 .0 L

⎟⎠

⎞

⎝

⎥⎦

⎤

⎢⎣

⎡⎟⎠

⎞ ⎝

+

Seismic Design:y

y

b r F

E 086 .0 L

⎟⎠

⎞

⎝

≤


Recognize that buckling o f a compression member is

non-ductile

Pcr

P

Pcr

δ

δ

P

Experimental Behavior of Brace Under Cyclic Axial LoadingExperimental Behavior of Brace Under Cyclic Axial Loading

δ

P

W6x20 Kl/r = 80

How Do We Achieve Ductile Response in Steel Structures ?

• Ductile limit states must precede brittle limit states

Ductile elements must be the weakest in the load path

Stronger is not better in ductile elements

Define Required Strength for brittle limit states based on

expected yield capacity of ductile element

• Avo id h igh str engt h st eels in du cti le elements

• Use cross-sections with low b/t ratios

• Provide adequate lateral bracing

• Recognize that compression member buckling is no n-ductile

• Provide connections that are stronger than members

How Do We Achieve Ductile Response in Steel Structures ?

Steel Ductility

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